MATH 1230 Homework 5.5, partial solutions
1. For each function f below, find the derivative f 0 . (You do NOT need to simplify.)
r
cos x
(a) f (x) =
x
f 0 (x) =
1 cos x −1/2 (− sin x)(x) − cos x
·
2
x
x2
(b) f (x) = [(6x + x5 )−1 + x]2
f 0 (x) = 2((6x + x5 )−1 + x) · (−(6x + x5 )−2 · (6 + 5x4 ) + 1)
(c) f (x) = sin(x + sin(x + sin(x)))
f 0 (x) = cos(x + sin(x + sin(x))) · (1 + cos(x + sin x) · (1 + cos x))
3. Find the equation of the line passing through (1, 2) that is tangent to the curve defined
by the relation 2x3 + 2y 3 = 9xy.
The point (1, 2) is on the curve, since 2(1)3 + 2(2)3 = 18 = 9(1)(2). So it
remains to find the slope of this tangent line at that point. To find y 0 , we
use implicit differentiation:
6x2 + 2y 2 · y 0 = 9y + 9xy 0
At the point (1, 2) this becomes 6 + 8y 0 = 18 + 9y 0 , and thus y 0 = −12.
Therefore, the equation of the line is y − 2 = −12(x − 1).
4. Evaluate the following limits.
sin 4x
(a) lim
x→0 sin 6x
sin x
= 1.
x→0 x
We use the special limit lim
sin 4x
sin 4x 6x 4
sin 4x
6x
2
2
2
= lim
·
· = lim
·
· =1·1· =
x→0 sin 6x
x→0 sin 6x 4x 6
x→0 4x
sin 6x 3
3
3
lim
Comment: Notice that we are using slightly modified versions of the above
‘special limit,’ such as limx→0 sin4x4x = 1. Why is this OK? Here’s an exercise
that show’s it’s OK: Let a > 0. If lim f (x) = L, then lim f (ax) = L. (Pf.
x→0
x→0
Let ε > 0. Since lim f (x) = L, there exists δ1 > 0 such that 0 < |x| < δ1 ⇒
x→0
|f (x) − L| < ε. Choose δ = δ1 /a. If 0 < |x| < δ, then 0 < |ax| < δ1 and hence
|f (x) − L| < ε, as required.)
University of Manitoba
Department of Mathematics
D. Krepski
Fall 2015
MATH 1230 Homework 5.5, partial solutions
sin(x − 1)
x→1 x2 + x − 2
(b) lim
lim
x→1
sin(x − 1)
1
1
1
sin(x − 1)
= lim
·
=1· =
2
x→1
x +x−2
(x − 1)
(x + 2)
3
3
5. In some applications (e.g. in physics) one uses the approximation sin x ≈ x for x small.
Explain why this approximation is justified (for x sufficiently small).
This is justified from the knowledge of (or meaning of) the limit
sin x
lim
= 1. For values of x that are sufficiently small (i.e. close to
x→0 x
0), sinx x ≈ 1, or sin x ≈ x for those values.
6. Suppose f and g are twice differentiable (i.e. second derivative exists). Show that
d2
(f (x)g(x)) = f 00 (x)g(x) + 2f 0 (x)g 0 (x) + f (x)g 00 (x).
2
dx
Differentiate the product rule formula to obtain the given formula. That
is, differentiate both sides of the formula
(f g)0 = f 0 g + f g 0
to get
(f g)00 = (f 0 g + f g 0 )0
= (f 0 g)0 + (f g 0 )0
= f 00 g + f 0 g 0 + f 0 g 0 + f g 00
= f 00 g + 2f 0 g 0 + f g 00
as required.
University of Manitoba
Department of Mathematics
D. Krepski
Fall 2015